The Density of an Iron Nucleus

solved NuclearSize

Q. Given the mass of iron nucleus as 55.85 u $m = 55.85$ u and A = 56 $A = 56$ , and taking R₀ as 1.2 fm $R_{0} = 1.2$ fm, what is the nuclear density, in kg/m³?

Inputs

mass

m

55.85 u

nucleons

A

size

R_{0}

1.2 fm

Outputs

density

ρ

How it's solved

Solved with NuclearSize

Given

$m = 55.85$ u
$A = 56$
$R_{0} = 1.2$ fm

Find $ρ$ .

Working

$m = 55.85$ u $= 9.27411069 \times 10^{−26}$ kg
$R_{0} = 1.2$ fm $= 1.2 \times 10^{−15}$ m
radius grows as the cube root(iron) $R = R_{0} \cdot A^{0.333333333}$
$R = 1.2 \times 10^{−15} \cdot 56^{0.333333333}$
$R = 4.59103484 \times 10^{−15}$ m
density of a nucleus(iron) $ρ = \frac{m}{\frac{4 \cdot π \cdot R^{3}}{3}}$
$ρ = \frac{9.27411069 \times 10^{−26}}{\frac{4 \cdot 3.14159265 \cdot {(4.59103484 \times 10^{−15})}^{3}}{3}}$
$ρ = 2.288 \times 10^{17}$ kg/m³
$ρ \approx 2.29 \times 10^{17}$ kg/m³to 3 significant figures

Conclusion

A nucleus's volume grows with the number of nucleons, so every nucleus has about the same enormous density, whatever the element.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :the_density_of_an_iron_nucleus do4  kind :numerical5  describes "Given the mass of iron nucleus as %{mass} and A = %{nucleons}, and taking R₀ as %{size}, %{density}, in " \6            "kg/m³?"7  given mass: "55.85 u"8  given nucleons: "56"9  given size: "1.2 fm"10  asks density: "what is the nuclear density"11end

solution.rb

1require_relative "../questions/the_density_of_an_iron_nucleus"2require_relative "../science/nuclei/nuclear_size"34solve :the_density_of_an_iron_nucleus, using: NuclearSize do5  objects :iron67  take mass: iron.mass8  take nucleons: iron.mass_number9  take size: R_01011  density = iron.nuclear_density12  answer density:, in: "kg/m3", figures: 31314  conclusion "A nucleus's volume grows with the number of nucleons, so every nucleus has about the same " \15             "enormous density, whatever the element."16end

playground.rb

1require_relative "../solutions/the_density_of_an_iron_nucleus"23play_with :the_density_of_an_iron_nucleus do4  title "The Density of an Iron Nucleus"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	NuclearSize
coulomb constant	$k$	N m² C⁻²	—	NuclearSize
elementary charge	$e$	C	—	NuclearSize
size of a nucleon s share	$R_{0}$	m	—	NuclearSize
mass number	$A$	—	0 … ∞	NuclearSize
atomic number	$Z$	—	0 … ∞	NuclearSize
mass	$m$	kg	0 … ∞	NuclearSize
nuclear radius	$R$	m	—	NuclearSize
nuclear density	$ρ$	kg/m³	—	NuclearSize
coulomb barrier	$U_{b}$	J	—	NuclearSize
nuclear radius of ground	$R_{g}$	m	—
mass number of ground	$A_{g}$	—	0 … ∞
nuclear radius of iron	$R$	m	—
mass number of iron	$A$	—	0 … ∞
nuclear density of ground	$ρ_{g}$	kg/m³	—
mass of ground	$m_{g}$	kg	0 … ∞
nuclear density of iron	$ρ$	kg/m³	—
mass of iron	$m$	kg	0 … ∞
coulomb barrier of ground	$U_{b,g}$	J	—
atomic number of ground	$Z_{g}$	—	0 … ∞
coulomb barrier of iron	$U_{b}$	J	—
atomic number of iron	$Z$	—	0 … ∞

R_{n} = R_{0} \cdot {A_{n}}^{0.333333333}

radius grows as the cube root

ρ_{n} = \frac{m_{n}}{\frac{4 \cdot π \cdot {R_{n}}^{3}}{3}}

density of a nucleus

U_{b,n} = \frac{k \cdot {(Z_{n} \cdot e)}^{2}}{2 \cdot R_{n}}

two alike nuclei just touching

R_{g} = R_{0} \cdot {A_{g}}^{0.333333333}

radius grows as the cube root(ground)

R = R_{0} \cdot A^{0.333333333}

radius grows as the cube root(iron)

ρ_{g} = \frac{m_{g}}{\frac{4 \cdot π \cdot {R_{g}}^{3}}{3}}

density of a nucleus(ground)

ρ = \frac{m}{\frac{4 \cdot π \cdot R^{3}}{3}}

density of a nucleus(iron)

U_{b,g} = \frac{k \cdot {(Z_{g} \cdot e)}^{2}}{2 \cdot R_{g}}

two alike nuclei just touching(ground)

U_{b} = \frac{k \cdot {(Z \cdot e)}^{2}}{2 \cdot R}

two alike nuclei just touching(iron)